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The fugitive is at the origin. They move at a speed of 1. There's a guard at every integer coordinate except the origin. A guard's speed is 1/100. The fugitive and the guards move simultaneously and continuously. At any moment, the guards only move towards the current position of the fugitive, i.e. a guard's trajectory is a pursuit curve. If they're within 1/100 distance from a guard, the fugitive is caught. The game is played on $\mathbb{R}^2$.

enter image description here

Can the fugitive avoid capture forever?


What I know:

  1. The distance between two guards is always non-increasing, but the farther away from the fugitive, the slower that distance decreases.

  2. If the fugitive runs along a straight line, they will always be caught by some far away guard. For example, if the fugitive starts at $(0.5, 0)$ and runs due north, they will be caught by a guard at about $(0, \frac{50^{100}}{4})$. Consult radiodrome for calculation.

  3. If there're just 2 guards at distance 1 from each other, then the fugitive can always find a path to safely pass between them. This is true regardless of the pair's distance and relative positions to the fugitive.

  4. The fugitive can escape if they're just "enclosed" by guards who're at distance 1 from each other:

enter image description here

The shape of the fence doesn't have to be rectangular. Circles or other shapes don't prevent escape either, regardless of the size.


3 and 4 are nontrivial, but can be proved by geometry argument alone without calculus. To avoid unnecessarily clustering the question, further details are given as an answer below for those who're interested, hopefully instrumental in solving the original problem.

Eric
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    @Stacker: Sure they'll "close in." But what's to stop the fugitive from escaping and leaving the close guards behind, always getting to a new region where the guards are still widely separated? My guess is that there is some speed where the fugitive can always escape. I wonder what that minimum speed is. – David G. Stork May 05 '21 at 03:25
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    Closely related by same poster: https://puzzling.stackexchange.com/questions/109739/can-you-survive-this-infinite-zombie-attack – Ross Millikan May 05 '21 at 03:46
  • If I had to guess, I'd say the fugitive is getting caught if there's an infinite number of guards since guards that are incredibly far away will be closing in on *basically* the origin, meaning the gap for escaping at that distance ("incredibly far away"), is closing in. Unless there's some certain sized and type of shape that can be made at a fast enough speed, I feel this fugitive will get caught. – Cotton Headed Ninnymuggins May 05 '21 at 03:53
  • @RossMillikan I feel that continuousness makes this different right? – Cotton Headed Ninnymuggins May 05 '21 at 03:58
  • @CottonHeadedNinnymuggins: I don't know. I just remembered the previous problem and thought it should be linked – Ross Millikan May 05 '21 at 04:00
  • Idea: we can prove that at some t > T, and beyond the distance from origin r > ct, there is a closed “ring” made by the guards to fully encircle the poor fugitive. (Or not) – Benjamin_Gal May 05 '21 at 04:54
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    My first intuition would be to try some "zig-zag" or "spiral-out" strategies. Otherwise, maybe a simpler problem could lead to some ideas. E.g. if we reduce the number of guards to a finite number and place every guard around a circle, then see [Escaping from a circle of fat lions.](https://math.stackexchange.com/q/1977972/318073). – Vepir May 05 '21 at 18:09
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    Is being able to escape from an arbitrarily large number of pursuers fine, or must it be infinite? – Caleb Briggs May 05 '21 at 20:36
  • @Vepir It seems that there is relation between the speed ratio and the number of lions - in other words - calculate the number of lions required to catch the fugitive and than it seems that there is a radius which will have enough guards. It seems to me that the fugitive will be caught after finite time. – Moti May 06 '21 at 02:08
  • @Vepir Zig-zagging would allow the fugitive to escape much much further than dashing in a straight line would. So I seriously doubt the straight line strategy in the circle of lion problem, especially if lions are slower than human. – Eric May 06 '21 at 03:26
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    If each guard can choose to stop, then they can always catch the fugitive, as [described in the comments here](https://www.reddit.com/r/math/comments/n58wii/interesting_question/). The idea is that a group of far-away guards can move towards the circumference of a large circle around the origin and stop there, forming a dense circular wall that the fugitive can’t pass through. The trouble with the original problem is that the guards are all constantly moving in the same direction, so they won’t naturally group together in a critical density that stops the fugitive from slipping though. – D G May 07 '21 at 00:49
  • @DG Thanks for the link! The problem gathers a lot of attention there and it's definitely an enjoyment to read all those helpful comments! If guards could strategize, they would simply form an impenetrable circle of sufficiently large radius before the fugitive get there. They could even form many such circles and shorten their catch time considerably, as shown in this [example](https://puzzling.stackexchange.com/questions/109820/catch-the-fugitive/109821#109821). – Eric May 07 '21 at 02:21
  • If the guards can pick direction of movement than it is trivial to create far away a circle with minimal density to catch the fugitive and than close on the fugitive. This is not the case here, which requires specific solution. – Moti May 08 '21 at 03:13
  • After further thought - moving along the diagonal with small circling around guards - one left and one right you avoid the guards along the diagonal, where the distance between them is not changing when they move toward the fugitive. The two guards closest to the one to be avoided, will not be able to catch the fugitive. – Moti May 09 '21 at 04:38
  • @Moti "The two guards closest to the one to be avoided, will not be able to catch the fugitive". Why not? Even if you remove all diagonal guards, the guards at (n, n+1) or (n+1, n) will catch you for sufficiently large n. ($n\gt 100^{100}$ will certainly do) – Eric May 09 '21 at 06:48
  • @CalebBriggs There's no difference. – Eric May 11 '21 at 13:26
  • Your game description is not complete: If a guard is at $(x+k,y+k)$ when the evader is at $x,y)$ does the guard move down or let? At the symmetric beginning it makes no difference, but afterwards, it does make a difference. – Mark Fischler May 11 '21 at 19:33
  • @MarkFischler Maybe you misunderstood the description. The game is not played on some lattice and guards don't move like rooks. Their speed vectors point at the current position of the fugitive. – Eric May 12 '21 at 00:36
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    [ideas](https://i.stack.imgur.com/zGc1q.gif) – martin May 20 '21 at 05:13
  • @martin That's amazing. Expert escaper :) What's the speed ratio? – Eric May 20 '21 at 05:27
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    @Eric 4:1. [It takes a while to get started though](https://i.stack.imgur.com/uG0xp.gif) – martin May 20 '21 at 05:45
  • @martin I was just about to ask about the unevenness of guards distribution. It's absolutely fascinating that guards distribution can be manipulated this way. – Eric May 20 '21 at 05:53
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    What would be the solution if the guards just move in a random uniform way in the plan? Another question is how close (forget the trivial solutions) the guards should be to guarantee they catch the fugitive? – Marco Bellocchi May 22 '21 at 10:35
  • @Cotton Headed Ninnimungis the continuity does not change anything, the difference is in the norm (euclidien here and the "norme 1" there – jcdornano May 23 '21 at 09:16
  • Maybe we can first ask ourself if there exist a catching distance (that is here 1/100) such that the fugitive can always escape. (I would say that it is the case, but i didnt thought enough) – jcdornano May 23 '21 at 09:53
  • @martin Did you control the fugitive manually or did they move according to some specified rules? – Eric May 25 '21 at 05:24
  • @Eric I used a discrete analogy to the pursuit curve whereby for every quarter of a unit step (or whatever ratio you choose) the pursuers make, the escapee makes a unit step to the point on the unit circle around the escapee that maximises the gap between itself and its pursuers. I modified my mathematica code from [here](https://math.stackexchange.com/questions/2427050/staff-icebreaker-is-stasis-ever-attained). It always eventually seems to fall into the pattern shown in my previous comment, despite starting point. Will post it if you like. – martin May 25 '21 at 08:59
  • @martin Interesting. Please do post it. I'm sure me and others would like to investigate further. – Eric May 25 '21 at 10:05
  • @Vepir I finally proved that the circle strategy fails. Take a look at my update. – Eric Jun 08 '21 at 16:17

3 Answers3

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As requested in the comments above, here is the mathematica code that generated this:

Might provide some insight. I used a discrete analog to the pursuit curve whereby for every quarter of a unit step the pursuers make (or whatever ratio you choose), the escapee makes a unit step to a point on the circumference of the unit circle around the escapee that maximises the gap between itself and its pursuers. Code modified from here.

Obviously the pursuer density increases with time. Whereas a straight line escape route plot of the distance between the escapee and it's pursuers is strictly decreasing

a plot of the distance between the escapee and it's pursuers for the local evasion strategy is, at first glance, not:

However, this is most likely misleading, since this discrete analog hides the continuous path that will at some point, no doubt pass within the restricted parameters:

martin
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    This is so cool. – K.defaoite May 26 '21 at 21:27
  • +1 for the insight. About accuracy, the simulation might be improved by using an adaptive step proportional to the distance to the closest pursuer. Ideally, it would also take into account the local curvature of the paths, not just the distance, but that can become complicated fast. – dxiv May 26 '21 at 21:46
  • @dxiv I did try an adaptive step, but that would imply a change in escapee speed, wouldn't it? If the escapee is allowed to accelerate, he will certainly escape. Re the accuracy, yes, this discrete analog is just a sketch really. – martin May 27 '21 at 04:22
  • @martin As long as all points use the same discretization step, the relative speeds would not change, and the outcome would not change. What would change, indeed, is the animation speed, which would appear to slow down and show more detail when points get close, or turn abruptly. P.S. Thanks for the pointer, unfortunately I won't have a chance to really play with it in the immediate future. – dxiv May 27 '21 at 04:43
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Proofs for points 3 and 4 in the question.

Proof for point 3

Let's denote the two guards and the fugitive by $G_1$, $G_2$ and $F$, respectively. If $\angle FG_1G_2\leq\pi/4$ at time $0$, then slipping through is possible because:

enter image description here

If $F$ travels to $G_1$ along the y axis, $G_2$ will always be within the eye-shaped intersection of the two circles before $F$ reaches $G_3$, because distance between any pair of guards is non-increasing. So the distance between $G_1$ and $G_2$ will be greater than $\sqrt2-1$ when $F$ reaches $G_3$. For a safety radius of 1/100 and speed ratio of 100, the fugitive can easily slip through between $G_1$ and $G_2$.

What about $\angle FG_1G_2\gt\pi/4$? Let's consider the extreme case where $\angle FG_1G_2=\pi/2$ at time $0$. Suppose $G_1(0)=(0,n)$ and $G_2(0)=(1,n)$. Notice if the fugitive runs leftward, the segment $G_1G_2$ will tilt counterclockwise, as illustrated below:

enter image description here

So if the fugitive runs clockwise along the arc of the circle of radius n centering at $G_1(0)$, $\angle FG_2G_1$ will be less than $\pi/4$ when the fugitive has traveled a distance of $n\pi/4\approx n$. Since the speed ratio is 100, the pair have traveled at most about $n/100$ and are still about $99n/100$ away, which means the distance between $G_1$ and $G_2$ has shrunk at most about 1% at this time. Now the fugitive can run straight for $G_2$ and slip through the pair as explained in the first paragraph.

In fact, there's even a better strategy! Notice $F$, $G_1$ and $G_2$ will be collinear sometime before the fugitive has traveled a distance of $n\pi/2\approx 2n$. By this time the pair of guards have traveled at most $n/50$ and are still about $49n/50$ away, which means distance between $G_1$ and $G_2$ has shrunk at most 2%. Since the triple are collinear, the gap between guards will remain constant at about 0.98 if the fugitive runs straight towards them. That's more than enough space for the fugitive to slip through at a leisurely pace. $\blacksquare$

Proof for point 4

Let's assume guards surround the fugitive with a circle of radius $R$. The fugitive moves straight toward some guard. When they're close to the guard, say at distance 1 away, the fugitive swerves to the right and passes every guard tangentially while keeping at about distance 1 away, as schematically illustrated below

enter image description here

Now, as the fugitive makes their clockwise movement, the distance between a pair of guards closest to them has to be no more than about 0.04 lest the target escapes. Since the distance between guards are non-increasing, when the fugitive finishes a full round, the distance between any two adjacent guards is no more than 0.04. But this is impossible given that the fugitive has moved no more than $R+2\pi R \approx 7R$. Because given the speed radio of 1:100, each guard has moved no more than $\frac{7}{100}R$. This means each guard is still at least $\frac{93}{100}R$ away from the origin. Hence the average distance between two adjacent guards is at least $\frac{93}{100}$. A contradiction. $\blacksquare$

Eric
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0

I think the fugitive can evade capture indefinitely. Here are my thoughts, which perhaps can be refined by someone else.

Consider a variation of the problem: the fugitive begins at (0,0) but there are approx $2\pi R$ guards uniformly-spaced around a circle centred at the origin of radius $R$, so that adjacent guards are spaced apart by ~1 unit. Also suppose that guards move toward the origin rather than toward the fugitive. (As a commenter observes, this is already basically the case for very distant guards.)

Suppose the fugitive runs at 100 units/s in a straight line. Then the prisoner intersects the circle of guards after approx $R/100$ seconds, at which point the circle of guards has approximate radius $\frac{99}{100} R$. The gap between adjacent guards is now 99/100 units, allowing the fugitive to easily slip through. Notice this quantity is independent of the initial radius $R$.

Now suppose there are circles of guards of radius 1, 2, 3, ... and on each circle, adjacent guards are separated by 1 unit. The distance between a pair of guards on different concentric circles is always at least 1, the fugitive can move freely in the gaps between concentric circles. Moreover, since the fugitive can slip past a circle of arbitrary radius, the fugitive can slip past all the circles and evade capture.

This concentric circle setup is interesting because the guards are initially distributed with roughly the same distribution and density as in the original problem. It is also still the case that at each point in time, the density of guards around the fugitive's position is increasing. But the fugitive has an easy escape strategy, and in fact, at no point in time is the fugitive even close to being caught.

It seems to me that arranging guards on concentric circles is essentially the same as arranging them on lattice points, since there's nothing special about integer coordinates. I also conjecture the analysis above doesn't meaningfully change if the guards move towards the prisoner rather than towards the origin.

The key idea in this problem is that the fugitive will be caught iff they become surrounded by a critical-density cloud of guards (i.e. a cloud of guards with more than one guard per ~0.01 square units). But the guards begin at a low density distribution, and they all move as one towards the fugitive, so their density increases very slowly; indeed, their density increases more slowly the further away they are from the fugitive. So the fugitive can always pass them before they've had time to accumulate sufficient density.

D G
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    I think there is a weak point in the argument, and that's the assumption that chasing the origin instead of the fugitive doesn't change the outcome. At first sight, the argument would unwind the same if the guards were placed on the $x = \pm 1/2$ verticals and the fugitive ran straight up along the $y$ axis. But in that case the fugitive would be eventually caught if the guards chased him, not the origin, and this would happen regardless of the relative speeds and sizes. – dxiv May 07 '21 at 07:28
  • @dxiv Let's say guards are evenly spaced on a very large circle centered at the origin. The radius of the circle is such that if the fugitive dash in a straight line between 2 guards, he will barely get caught, i.e. closest distance to a guard is 1/100. Can he do better than this by going along some other curve? You see in your =±1/2 example, when guards to the north is about to catch him, the fugitive can just turn left or right and continue escaping for a distance comparable to his northward dash, and then zig-zag again... – Eric May 07 '21 at 08:32
  • The difficulty in the escape as I see it is that if the a guard would be able to catch the fugitive along the $y$ axis, then you have quite a long line of guards already too close to slip through, so you really do need to run east or west for a similar distance before you might find a gap. But while you do that the guards to the north and south of this line are closing in on your new $y$ coordinate. Can you find a gap in the line to your north before the guards to the south block your path? – David K May 07 '21 at 12:10
  • @DavidK Yes, you can. The "closing in" effect is only prominent along the north-south direction. Play the simulation mentioned in the post and you'll see. – Eric May 07 '21 at 14:07
  • @Eric My comment was only meant to point out that the straight line escape does not work as proposed, and the reason why it cannot work. – dxiv May 07 '21 at 17:08
  • @Eric As for the question itself, that's harder to tell. Intuition can play tricks with the mind when infinities are involved, and simulations don't help all that much unless you can estimate upfront how far you need to go. It doesn't help to know that the first million steps succeeded, if the capture is expected after a trillion steps. – dxiv May 07 '21 at 17:15
  • @Eric I looked at the "simulation" page when it was first mentioned but all I found was a github page with some files documented in Chinese. I wasn't inclined to try to decipher it. Certainly as you run up the $y$ axis I would expect less "closing in" of guards that started along lines of constant $x$ than of guards that started along lines of constant $y$, but that doesn't answer the question of whether the gaps are still big enough by the time you've run the second leg of your escape in the perpendicular direction. – David K May 07 '21 at 18:00
  • Maybe some kind of argument with the stereographic projection of $1/100$-square around guards on some wise square around F, would help. The farest are the gaurd the smallest is the segment that is the projection of their square. The good size for the wise square would be such that he capture enough information that it will be enough for him to move according to it (maye be the "gaps" in the wise sqare. – jcdornano May 24 '21 at 10:10
  • I had this "steregraphic idea" in the resulution of another puzzle posted by Eric . The solution is here :(just look at the answer, the steregraphic argument is after the conclusion) https://math.stackexchange.com/questions/4147724/exemple-of-chassing-puzzles-idea-of-extending-from-the-way-to-solve-create-each/4147833#4147833. Also the game is discribed here : https://mathoverflow.net/questions/389872/can-the-fugitive-escape/392906#392906 – jcdornano May 24 '21 at 10:14